A particle of mass 0.300 kg is attached to the 100-cm mark of a meterstick of ma

Question

A particle of mass 0.300 kg is attached to the 100-cm mark of a meterstick of mass 0.175 kg. The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 2.00 rad/s.

(a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50.0-cm mark. The moment of inertia of several objects is the sum of the moments of inertia of the individual objects.

=________kg · m2/s

(b) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark.

=_________kg · m2/s

Explanation / Answer

a) mass of meterstick=0.175 kg

mass of particle=0.3 kg

distnace of mass from pivot point=0.5m

mass moment of inertia of system = 0.175*12/12 + 0.3*0.52 = 0.0896 kg-m2

angular velocity=2 rad/s

angu;ar momentum = 0.0896*2=0.179 kg-m2/s

b)mass moment of inertia of system =  0.175*12/12 + 0.175*0.52+0.3*12 =0.35833 kg-m2

angular momentum=0.35833*2=0.7167 kg-m2/s

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